A spherical ball of radius $1 \times 10^{-4} \mathrm{~m}$ and density $10^5 \mathrm{~kg} / \mathrm{m}^3$ falls freely under gravity through a distance $h$ before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately:

(The coefficient of viscosity of water is $9.8 \times 10^{-6} \mathrm{~N} \mathrm{~s} / \mathrm{m}^2$)

<p>To solve this problem, we can use the concepts of terminal velocity and the forces acting on the spherical ball. First, let's analyze the situation step-by-step.</p> <p>When the ball falls freely under gravity, it achieves a terminal velocity $v_t$ in water. This terminal velocity is reached when the gravitational force is balanced by the drag force and the buoyant force in the water.</p> <p>The forces acting on the ball are:</p> <p>1. Gravitational Force: $F_g = mg$</p> <p>2. Buoyant Force: $F_b = \rho_{\text{water}} V g$</p> <p>3. Drag Force: $F_d = 6 \pi \eta r v_t$</p> <p>Where,</p> <p>$m$ is the mass of the ball.</p> <p>$g$ is the acceleration due to gravity ($9.8 \, \mathrm{m/s^2}$).</p> <p>$\rho_{\text{water}}$ is the density of water ($1000 \, \mathrm{kg/m^3}$).</p> <p>$V$ is the volume of the ball ($\frac{4}{3} \pi r^3$).</p> <p>$\eta$ is the coefficient of viscosity of water ($9.8 \times 10^{-6} \, \mathrm{Ns/m^2}$).</p> <p>$r$ is the radius of the ball ($1 \times 10^{-4} \, \mathrm{m}$).</p> <p>$v_t$ is the terminal velocity.</p> <p>Using the equilibrium condition at terminal velocity:</p> <p>$F_g = F_b + F_d$</p> <p>$mg = \rho_{\text{water}} V g + 6 \pi \eta r v_t$</p> <p>First, compute the mass of the ball:</p> <p>$m = \rho_{\text{ball}} \times V = \rho_{\text{ball}} \times \frac{4}{3} \pi r^3$</p> <p>$$m = 10^5 \, \mathrm{kg/m^3} \times \frac{4}{3} \pi (1 \times 10^{-4} \, \mathrm{m})^3$$</p> <p>$m = 10^5 \,\mathrm{kg/m^3} \times \frac{4}{3} \pi \times 10^{-12} \,\mathrm{m^3}$</p> <p>$m = \frac{4}{3} \pi \times 10^{-7} \, \mathrm{kg}$</p> <p>Next, solve for the terminal velocity $v_t$ using the equilibrium equation:</p> <p>$mg = \rho_{\text{water}} \frac{4}{3} \pi r^3 g + 6 \pi \eta r v_t$</p> <p>$v_t = \frac{mg - \rho_{\text{water}} \frac{4}{3} \pi r^3 g}{6 \pi \eta r}$</p> <p>$$v_t = \frac{\frac{4}{3} \pi \times 10^{-7} \times 9.8 - 1000 \times \frac{4}{3} \pi (1 \times 10^{-4})^3 \times 9.8}{6 \pi \times 9.8 \times 10^{-6} \times 10^{-4}}$$</p> <p>$$v_t = \frac{\frac{4}{3} \pi \times 10^{-7} \times 9.8 - 1000 \times \frac{4}{3} \pi \times 10^{-12} \times 9.8}{6 \pi \times 9.8 \times 10^{-10}}$$</p> <p>$$v_t = \frac{\frac{4}{3} \pi \times 9.8 \times 10^{-7} (1 - 10^{-5})}{6 \pi \times 9.8 \times 10^{-10}}$$</p> <p>$v_t = \frac{\frac{4}{3} \times 10^{-7}}{6 \times 10^{-10}}$</p> <p>$v_t = \frac{4}{18} \times 10^3 \, \mathrm{m/s}$</p> <p>$v_t \approx 222.22 \, \mathrm{m/s}$</p> <p>The height $h$ required to reach this terminal velocity while the ball falls freely under gravity can be found using the kinematic equation:</p> <p>$v_t = \sqrt{2gh}$</p> <p>$h = \frac{v_t^2}{2g}$</p> <p>$h = \frac{(222.22)^2}{2 \times 9.8}$</p> <p>$h = \frac{49328.88}{19.6}$</p> <p>$h \approx 2517.8 \, \mathrm{m}$</p> <p>So, the closest value of $h$ is approximately:</p> <p><b>Option A: 2518 m</b>.</p>